Let the concatenation abc represent the last three digits of the 7th power of randomly chosen 4digit integer.
Evaluate the probability that at least one of the digits a,b, or c is zero.
Bonus (D4) task: Still addressing the last 3 digits, derive a formula for p(k,l) expressing the probability for the lth power of kdigit number, l must be prime in respect to 10 , and no less than k.
When working with the last 3 digits of a number, the value of k, the number of digits in the number, doesn't matter so long as k > 2, as we use modular arithmetic (mod 1000). What's would be more interesting would be varying the number of digits at the end, rather than limiting it to 3 digits. Call this d, for the last d digits.
For d = 2 To 6: lim = Int(10 ^ d + 0.5)
For p = 2 To 10
ct0 = 0: ctnz = 0
For i = 0 To lim  1
prod = i
For j = 2 To p
prod = prod * i
q = Int(prod / lim)
prod = prod  q * lim
Next
s$ = LTrim(Str(prod))
If InStr(s$, "0") > 0 Or Len(s$) < d Then ct0 = ct0 + 1 Else ctnz = ctnz + 1
Next
Text1.Text = Text1.Text & mform(ct0 / lim, "#0.000000"): DoEvents
Next: Text1.Text = Text1.Text & crlf
Next
l (power)
d 2 3 4 5 6 7 8 9 10
2 0.220000 0.180000 0.180000 0.200000 0.220000 0.180000 0.180000 0.180000 0.300000
3 0.328000 0.252000 0.260000 0.270000 0.292000 0.252000 0.260000 0.252000 0.380000
4 0.403600 0.326800 0.424000 0.345500 0.402800 0.329700 0.424000 0.329700 0.486000
5 0.464400 0.394870 0.481600 0.414300 0.457520 0.400480 0.481600 0.400480 0.532400
6 0.518276 0.455758 0.533472 0.472870 0.519268 0.462307 0.533440 0.462307 0.586660
When d is 3 and l is 3, 7 or 9, the probability is 0.252. The same holds for power 11, 13, 17, 19 and 21.
So a conjecture is that when using the last three digits and the power is relatively prime to 10, the probability is always 0.252. The number of digits, k, does not affect it so long as it is 3 or more, so that taking the last 3 digits is meaningful. Of course 2 is not relatively prime to 10 anyway.

Posted by Charlie
on 20141007 10:35:23 